Positive definite kernel vs. positive definite function What is the difference between positive definite kernels and positive definite functions?  As I understand it, a positive definite kernel is a positive definite function if it is translation invariant.  Is this correct?  If not, what is the actual relationship between these concepts?  I'm reading Learning with Kernels and using this text, Wikipedia, and Google, haven't been able to find a good answer to this question.
References are appreciated!
 A: Wendland's Scattered Data Approximation covers this topic in chapter 6. Below, I have paraphrased relevant definitions.
Definition 6.1 A continuous function $\Phi:\mathbb{R}^d\rightarrow \mathbb{C}$ is called positive definite on $\mathbb{R}^d$ if for all $N\in\mathbb{N}$, all pairwise distinct $X=\{x_1,\dots,x_N\}\subset\mathbb{R}^d$, and all $\alpha\in\mathbb{C}^N\backslash\{0\}$ we have
\begin{equation*}
\sum_{j=1}^N 
\sum_{k=1}^N 
\alpha_j
\bar{\alpha_k}
\Phi(x_j - x_k)
>0
\end{equation*} 
Definition 6.24 A continuous kernel $\Phi:\mathbb{R}^d\times\mathbb{R}^d\rightarrow \mathbb{C}$ is called positive definite on $\mathbb{R}^d$ if for all $N\in\mathbb{N}$, all pairwise distinct $X=\{x_1,\dots,x_N\}\subset\mathbb{R}^d$, and all $\alpha\in\mathbb{C}^N\backslash\{0\}$ we have
\begin{equation*}
\sum_{j=1}^N 
\sum_{k=1}^N 
\alpha_j
\bar{\alpha_k}
\Phi(x_j,x_k)
>0
\end{equation*} 
To answer the posed question, functions are basically translation invariant kernels.
Suppose $\Phi:\mathbb{R}^d\times\mathbb{R}^d\rightarrow \mathbb{C}$ is translation invariant, i.e. $\Phi(x-t,y-t)=\Phi(x,y)$ for all $t,x,y$. Then $\Phi(x,y)=\Phi(x-y,0)$, so $\Phi$ is completely determined by $\Phi(z,0)$ for $z\in \mathbb{R}^d$. Hence, we can view it as a function on $\mathbb{R}^d$.
Conversely, every function gives rise to a kernel that is translation invariant.
